Calculating work heat, and efficiency given a TS diagram

[abobo37]

Homework Statement

Is it possible to calculate the work done heat transfer, and efficiency of an object in a thermodynamic system, given the Temperature vs Entropy graph?

Example:

Homework Equations

∆U=Q-W
∆S=dQ/T
e=1-(Qc/Qh)
e=W/Qh

The Attempt at a Solution

Since this is a T vs S diagram,i can understand how the product(area) will be heat.
Since it's a cyclic process, I can also understand how Q=W , since ∆U=0
Currently I'm thinking that the area inside the triangle will be W, which also corresponds to Q. However, I could barely find any documentation on this, which is why I wanted to confirm it.

As for efficiency, we are given two equations:
1-(Qc/Qh)
and
W/Qh
Would both of them be correct here, and would Qh and Qc be T2 and T1 respectively?

Thanks!

 

[SteamKing]

Homework Statement

Is it possible to calculate the work done heat transfer, and efficiency of an object in a thermodynamic system, given the Temperature vs Pressure graph?

Example:

Homework Equations

∆U=Q-W
∆S=dQ/T
e=1-(Qc/Qh)
e=W/Qh

The Attempt at a Solution

Since this is a T vs S diagram,i can understand how the product(area) will be heat.
Since it's a cyclic process, I can also understand how Q=W , since ∆U=0
Currently I'm thinking that the area inside the triangle will be W, which also corresponds to Q. However, I could barely find any documentation on this, which is why I wanted to confirm it.

As for efficiency, we are given two equations:
1-(Qc/Qh)
and
W/Qh
Would both of them be correct here, and would Qh and Qc be T2 and T1 respectively?

Thanks!

Is a T-S diagram a plot of Temperature vs. Pressure?

 

[abobo37]

Is a T-S diagram a plot of Temperature vs. Pressure?

I'm so sorry, I meant entropy. I will edit it! :D

 

[rude man]

Homework Statement

3. The Attempt at a Solution [/B]

Since this is a T vs S diagram,i can understand how the product(area) will be heat.
Since it's a cyclic process, I can also understand how Q=W , since ∆U=0
Currently I'm thinking that the area inside the triangle will be W, which also corresponds to Q. However, I could barely find any documentation on this, which is why I wanted to confirm it.

As for efficiency, we are given two equations:
1-(Qc/Qh)
and
W/Qh
Would both of them be correct here

Certainly, since these are both statements of the 1st law.

and would Qh and Qc be T2 and T1 respectively?

For a Carnot cyccle, yes. A T-S diagram is valid for reversible processes only since entropy is defined for equilibrium states only.
T-S diagrams are ideal for reading heats in a closed cycle:
Qh = positive area under curve (going from left to right)
Qc = negative area under curve (going from right to left)
So Qh - Qc per cycle = area within the closed curve, as you correctly state, and = W also. And e = 1 - (Qc/Qh).

 

[abobo37]

Thanks for confirming!